Question

Notify Me Bookmark

A common-source amplifier is shown. The transistor parameters are VTN = −0,8 V, Kn = 2 mA/V2, and λ = 0. The circuit parameters are VDD = 3,3 V and RL = 10 kΩ. a) Design the circuit such that IDQ = 0,5 mA and VDSQ = 2 V. b) Determine the small-signal voltage gain. c) Draw the load line sketch. Calculate the transition point, VDS(sat) on the AC load line. d) Calculate the output swing around the DC operating point (upwards and downwards). Note that δvo is required, and not δvDS!

A common-source amplifier is shown. The transistor parameters are VTN = −0,8 V, Kn = 2 mA/V2, and λ = 0. The circuit parameters are VDD = 3,3 V and RL = 10 kΩ. a) Design the circuit such that IDQ = 0,5 mA and VDSQ = 2 V. b) Determine the small-signal voltage gain. c) Draw the load line sketch. Calculate the transition point, VDS(sat) on the AC load line. d) Calculate the output swing around the DC operating point (upwards and downwards). Note that δvo is required, and not δvDS!

Image text
A common-source amplifier is shown. The transistor parameters are V T N = − 0 , 8 V , K n = 2 m A / V 2 , and λ = 0 . The circuit parameters are V D D = 3 , 3 V and R L = 10 k Ω .
a) Design the circuit such that I D Q = 0 , 5 m A and V D S Q = 2 V .
b) Determine the small-signal voltage gain.
c) Draw the load line sketch. Calculate the transition point, V D S ( s a t ) on the AC load line.
d) Calculate the output swing around the DC operating point (upwards and downwards). Note that δ v o is required, and not δ v D S !

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

4.22 For the common-source amplifier in Figure P4.22, the transistor parameters are VTN = −0.8 V, Kn = 2 mA/V2, and λ = 0. The circuit parameters are VDD = 3.3 V and RL = 10 kΩ. (a) Design the circuit such that IDQ = 0.5 mA and VDSQ = 2 V. (b) Determine the small-signal voltage gain. Figure P4.22
Solution
7 0

The transistor parameters are VTN = −0.7 V, Kn = 25 mA/V2 and λ = 0. a) Find IDQ and VDSQ and determine the small-signal voltage gain. b) Draw the load line sketch. Calculate the transition point VDS (sat) on the AC load line. c) Calculate the maximum undistorted output swing Δvo(max).

Solution
0 0

Ex 4.2: For the circuit shown in Figure 4.1, VDD = 3.3 V and RD = 10 kΩ. The transistor parameters are VTN = 0.4 V, kn′ = 100 μA/V2, W/L = 50, and λ = 0.025 V−1. Assume the transistor is biased such that IDQ = 0.25 mA. (a) Verify that the transistor is biased in the saturation region. (b) Determine the small-signal parameters gm and ro. (c) Determine the small-signal voltage gain. (Ans. (a) VGSQ = 0.716 V and VDSQ = 0.8 V so that VDS > VDS (sat); (b) gm = 1.58 mA/V, ro = 160 kΩ; (c) -14.9) Figure 4.1 NMOS common-source circuit with time-varying signal source in series with gate dc source

Ex 4.2: For the circuit shown in Figure 4.1, VDD = 3.3 V and RD = 10 kΩ. The transistor parameters are VTN = 0.4 V, kn′ = 100 μA/V2, W/L = 50, and λ = 0.025 V−1. Assume the transistor is biased such that IDQ = 0.25 mA. (a) Verify that the transistor is biased in the saturation region. (b) Determine the small-signal parameters gm and ro. (c) Determine the small-signal voltage gain. (Ans. (a) VGSQ = 0.716 V and VDSQ = 0.8 V so that VDS > VDS (sat); (b) gm = 1.58 mA/V, ro = 160 kΩ; (c) -14.9) Figure 4.1 NMOS common-source circuit with time-varying signal source in series with gate dc source
Solution
0 0